Paper stories

# Return the desired percentiles plus the geometric mean
bp.vals <- function(x, probs=c(.05, 0.25, 0.5, 0.75, .95), width=0.8) {
  r <- quantile(x, probs=probs , na.rm=TRUE)
  r <- c(r, width)
  names(r) <- c("ymin", "lower", "middle", "upper", "ymax", "width")
  r
}

\(I_o\) for all itineraries

ggplot(data = user_itineraries_complete, 
       aes(x = trip_length_bucket, 
           y = io)) +
    stat_summary(fun.data = bp.vals, geom = "boxplot", fill ="#fc8d62") +
    scale_y_log10(breaks = c(1:5)) +
    coord_flip() +
    xlab(expression(paste("d"[o], "(u) in mins"))) +
    ylab(expression(paste("User choice inefficiency (i"[o], ")"))) +
    guides(fill = FALSE)
ggsave("io-for-all.pdf", width = 5, height = 2.2)

Overall, longer itineraries tend to incurr in higher inefficiencies. For 20-30min itineraries, more than half have no inefficiency. For 40-50 min itineraries, half of the itineraries have an inefficiency of 1.6 or higher. For a 45 mins itinerary, \(i_o = 1.6\) implies there was an 28-min alterantive.

\(i_s\) for all \(i_c\)

ggplot(data = user_itineraries_complete, 
       aes(x = trip_length_bucket, 
           y = is)) +
    stat_summary(fun.data = bp.vals, geom = "boxplot", fill ="#7DBBC3") +
    scale_y_log10(breaks = c(1, 1.2, 1.4)) +
    coord_flip() +
    xlab(expression(paste("d"[o], "(u) in mins"))) +
    ylab(expression(paste("System operation inefficiency (i"[s], ")"))) +
    guides(fill = FALSE)
ggsave("is-for-all.pdf", width = 5, height = 2.2)

Contrary to what is experienced by user, system efficiency is concentrated on shorter itineraries. For 40-50-min itineraries, over 75% of all itineraries have no inefficiency.

It seems that most inefficiency in longer itineraries experienced by users is not due to system operation inefficiency.

Considering \(i_c\)

How often do users choose the best itinerary according to the schedule?

user_itineraries_complete %>% 
    group_by(trip_length_bucket) %>% 
    select(ic) %>% 
    summarise(best = sum(ic == 0, na.rm = T) / sum(!is.na(ic)), 
              n = sum(!is.na(ic))) 

Adding missing grouping variables: `trip_length_bucket`

Most of the time we can identify which trip from the schedule a user took, the user took the best according to the schedule. This happens 85-89% of the itineraries, depending on their duration.

Did the schedule happen faithfully when the user followed it?

user_itineraries_complete %>% 
    group_by(trip_length_bucket) %>% 
    filter(!is.na(ic), ic == 0) %>% 
    ggplot(aes(x = trip_length_bucket, 
               y = is)) +
    stat_summary(fun.data = bp.vals, geom = "boxplot", fill ="#7DBBC3") +
    # scale_y_log10() +
    coord_flip() +
    xlab(expression(paste("d"[o], "(u) in mins"))) +
    ylab(expression(paste("System operation inefficiency (i"[s], ")"))) +
    guides(fill = FALSE)

     
user_itineraries_complete %>% 
    group_by(trip_length_bucket) %>% 
    filter(!is.na(ic), ic == 0) %>% 
    ggplot(aes(x = trip_length_bucket, 
               y = io)) +
    stat_summary(fun.data = bp.vals, geom = "boxplot", fill ="#fc8d62") +
    scale_y_log10(breaks = c(1, 1.1, 1.2, 1.3)) +
    coord_flip() +
    xlab(expression(paste("d"[o], "(u) in mins"))) +
    ylab(expression(paste("User choice inefficiency (i"[o], ")"))) +
    guides(fill = FALSE)
ggsave("io-for-ic_eq_zero.pdf", width = 5, height = 2.2)

Following the schedule in general leads users to a low inefficiency (\(i_o = 1\) for 66-80% of the itineraries) longer itineraries tend to have have lower inefficinecy.

\(i_c > 0\)

$i_c > 0 $

plotly::ggplotly(p)

We recommend that you use the dev version of ggplot2 with `ggplotly()`
Install it with: `devtools::install_github('hadley/ggplot2')`

When the user deviates from the itinerary recommended by the schedule, the user outperforms the choice proposed by the schedule a significant proportion of the time.